Q. A potentiometer wire, 10 m long, has a resistance of 40$\Omega$. It is connected in series with a resistance box and a 2V storage cell. If the potential gradient along the wire is $\left( 0.1\ mVcm^{- 1} \right)$, the resistance unplugged in the box is
A) 260$\ \Omega$
B) 760$\ \Omega$
C) 960$\ \Omega$
D) 1060$\ \Omega$
Explanation
(b) Potential gradient along wire
$= \frac{potential\ difference\ along\ wire}{length\ of\ wire}$
$\therefore 0.1 \times 10^{- 3} = \frac{I \times 40}{1000}Vcm^{- 1}$
Current in wire.
$I = \frac{1}{400}A\ \ \ \ or\ I = \frac{E}{R + R^{'}}\ $
$\therefore\ \frac{2}{40 + R} = \frac{I}{400}$
Or $R = 800\ - \ 40\ = 760\Omega$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.